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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>To convert (<a href="" class="xref" data-knowl="./knowl/eq5_8.html" title="Equation 5.4.1">(5.4.1)</a>) into a constant coefficient equation, we introduce a change of variables</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_8.html">
\begin{equation*}
x=e^{z} \to z=\ln x.
\end{equation*}
</div>
<p class="continuation">Then,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_8.html">
\begin{equation*}
\begin{aligned}
&amp; \frac{\textrm{d} y}{\textrm{d} x}=\frac{\textrm{d} y}{\textrm{d} z}\cdot \frac{\textrm{d} z}{\textrm{d} x}=\frac{1}{x} \frac{\textrm{d} y}{\textrm{d} z},\\
&amp;\frac{\textrm{d}^2 y}{\textrm{d} x^2}=\frac{\textrm{d}}{\textrm{d} x} \left( \frac{1}{x}\frac{\textrm{d} y}{\textrm{d} z} \right)=-\frac{1}{x^2} \frac{\textrm{d} y}{\textrm{d} z}+\frac{1}{x} \frac{\textrm{d}}{\textrm{d} x} \left( \frac{\textrm{d} y}{\textrm{d} z}\right)=-\frac{1}{x^2} \frac{\textrm{d} y}{\textrm{d} z}+\frac{1}{x^2} \frac{\textrm{d}^2 y}{\textrm{d} z^2} .
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Substituting above equations into the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_8.html">
\begin{equation}
-\frac{\textrm{d} y}{\textrm{d} z}+\frac{\textrm{d}^2 y}{\textrm{d} z^2}+\alpha \frac{\textrm{d} y}{\textrm{d} z}+\beta y=0 \to \frac{\textrm{d}^2 y}{\textrm{d} z^2}+(\alpha-1) \frac{\textrm{d} y}{\textrm{d} z}+\beta y=0.\tag{5.4.2}
\end{equation}
</div>
<span class="incontext"><a href="sec5_4.html#p-227" class="internal">in-context</a></span>
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